In definition 1.1 of these notes on factorization systems, (III) calls the factorization functorial if given the solid diagram described, there's a unique horizontal arrow making both squares commute. Why does this deserve the name 'functorial'?


1 Answer 1


Let $[n]$ denote the category $0\to 1\to\cdots\to n$. Given a factorization system $(L, R)$ in a category $\mathcal{C}$, we can choose a map $F_0\colon\operatorname{Ob}\mathcal{C}^{[1]}\to\operatorname{Ob}\mathcal{C}^{[2]}$ which maps an object $x\xrightarrow{f} y$ in $\operatorname{Ob}\mathcal{C}^{[1]}$ to an object $x\xrightarrow{e} E\xrightarrow{m} y$ in $\mathcal{C}^{[2]}$ such that $e\in L$, $m\in R$ and $f=me$.

If your factorization system is functorial, this extends to a functor $F\colon\mathcal{C}^{[1]}\to\mathcal{C}^{[2]}$ mapping a morphism $(u,v)$ in $\mathcal{C}^{[1]}$ to a morphism $(u,\omega,v)$ in $\mathcal{C}^{[2]}$ where $\omega$ is the unique morphism required in (III), making the respective diagram commutative. The uniqueness guaranties that $F$ is indeed a functor, i.e. that given two morphism $(u,v),(u',v')\in\mathcal{C}^{[1]}$, we have

$$ F((u,v))\circ F(u',v') = (u,\omega,v)\circ (u',\omega',v') = (u\circ v',\omega\circ\omega',v\circ v') = F((u\circ u',v\circ v'))$$

Also note that even though the choice of $F_0$ is not unique, (III) guarantees that it is unique up to isomorphisms. To prove this, take two factorizations $(e, m)$ and $(e',m')$ of the same morphism $f$ and consider the diagrams $$ \require{AMScd} \begin{CD} x@>\operatorname{id}>> x @>\operatorname{id}>> x\\ @V{e}VV @V{e'}VV @V{e}VV \\ E@>\omega>>E' @>\omega'>> E \\ @V{m}VV @V{m'}VV @V{m}VV\\ y@>\operatorname{id}>> y @>\operatorname{id}>> y \end{CD} $$ and $$ \begin{CD} x@>\operatorname{id}>> x\\ @V{e}VV @V{e}VV \\ E@>\omega'\circ\omega>> E \\ @V{m}VV @V{m}VV\\ y@>\operatorname{id}>> y \end{CD} $$ By (III): $\omega'\circ\omega$ is unique, hence $\omega'\circ\omega = \operatorname{id}$. Thus our factorization functor $F$ is determined up to isomorphisms, and it makes sense to talk about the associated factorization functor.

  • $\begingroup$ Quick question - in the last sentence you say $F$ is determined up to isomorphism, but it seems $F$ is only determined up to isomorphism object-wise, i.e $F(u,v)$ is determined up to iso for each object $(u,v)$. Why is it true that all extensions are naturally isomorphic as functors? $\endgroup$
    – Arrow
    Apr 26, 2016 at 15:39
  • $\begingroup$ @Arrow I thin you mixed something up here. $F_0$ is uniquely determined up to isomorphisms and only defined on objects $x\xrightarrow{f} y$ in $\mathcal{C}^{[1]}$. The extension to a functor $F$ is then given by mapping morphisms $(u,v)$ in $\mathcal{C}^{[1]}$ to morphisms $(u,\omega, v)$. I'm not sure right now if this extension is unique, or if you could do something weird like sending all morphisms in $\mathbf{Grp}^{[1]}$ to those mapping everything to the neutral element. But it's at least the unique natural extension. Where uniqueness is guaranteed by (III). $\endgroup$
    – roman
    Apr 28, 2016 at 8:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .